A-Level Mathematics October/November 2024 Q9(a): The equation of a curve is y = 4+5x+6x² – 3x³. Find the set of values of x for which y…

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  The equation of the curve is $y = 4+5x+6x^2 – 3x^3$.

  A function $y$ decreases as $x$ increases when its derivative, $\frac{dy}{dx}$, is negative.

  First, we find the derivative of the function: $y = 4+5x+6x^2 – 3x^3$ $\frac{dy}{dx} = 5 + 2(6)x - 3(3)x^2$ $\frac{dy}{dx} = 5 + 12x - 9x^2$

  For the function to be decreasing, we require $\frac{dy}{dx} < 0$. $5 + 12x - 9x^2 < 0$

  To solve this quadratic inequality, we first find the critical values by solving the corresponding equation: $5 + 12x - 9x^2 = 0$ Rearranging for easier factorization: $9x^2 - 12x - 5 = 0$

  We can factorize the quadratic: $(3x+1)(3x-5) = 0$

  The critical values are: $3x+1 = 0 \implies x = -\frac{1}{3}$ $3x-5 = 0 \implies x = \frac{5}{3}$

  The quadratic $5 + 12x - 9x^2$ has a negative coefficient for $x^2$, so it is a downward-opening parabola (n-shaped). The expression is less than zero ($<0$) outside the roots.

  Therefore, the set of values of $x$ for which $y$ is decreasing is: $x < -\frac{1}{3}$ or $x > \frac{5}{3}$

  Final Answer: $x < -\frac{1}{3}, x > \frac{5}{3}$

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  How the marks are awarded

  • B1 — Correctly differentiating the given cubic equation to obtain the quadratic expression $5+12x-9x^2$.
  • M1 — Setting their derivative to be less than zero (or equal to zero) and attempting to find the two critical values by solving the resulting quadratic equation, e.g., by factorisation or using the quadratic formula.
  • A1 — Obtaining the correct critical values of $x = -1/3$ and $x = 5/3$ from the correct quadratic equation.
  • A1 FT — Stating the correct final set of values, $x < -1/3$ and $x > 5/3$. This mark is for the correct inequality based on their critical values and the shape of the quadratic graph. FT (Follow Through) is allowed from their critical values.

  Common mistakes

  • Solving the wrong inequality, i.e., finding where the function is increasing (rac{dy}{dx} > 0) instead of decreasing (rac{dy}{dx} < 0).
  • Incorrectly determining the inequality region after finding the critical values. A common error is writing $-1/3 < x < 5/3$, which would be the solution for an upward-opening parabola.
  • Making a sign error when rearranging the quadratic, for example, changing $9x^2 - 12x - 5 = 0$ to $(3x-1)(3x+5)=0$.
  • Using the word 'and' to connect the two parts of the solution, e.g., '$x < -1/3$ and $x > 5/3$', which is a logical impossibility. A comma or the word 'or' should be used.

  Examiner tip: To solve a quadratic inequality, always find the critical values first and then use a quick sketch of the parabola to determine whether the solution lies between or outside these values.

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